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Mathematics — Standard (041) — CBSE Class 10 board question

Q1. [3] § 8.2 Trigonometric Ratios § 8.4 Trigonometric Identities
Prove that: $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$
Previously asked in CBSE board exam
2025 30/1/1 Q27 (OR-2)
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer

To Prove: $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$

Proof:

$$\text{LHS} = \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta}$$

Taking LCM as $\sin\theta(1 + \cos\theta)$:

$$= \frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$$

$$= \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)}$$

$$= \frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$$

Using identity $\sin^2\theta + \cos^2\theta = 1$:

$$= \frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} = \frac{2 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$$

$$= \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \frac{2}{\sin\theta} = \text{RHS} \quad \blacksquare$$

Source: Chapter 8, Section 8.4 Trigonometric Identities

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Explanation
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